Differential Equation

Main Question or Discussion Point

dy/dx = (3x^2)(e^-y)

1/(e^-y) dy = 3x^2 dx
1/(e^-y) = x^3 + C

Can anybody please give me any help or advice on to where to go from here? I'm having a really tough time with this one.
 

Answers and Replies

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[tex]\frac{dy}{dx}=3x^{2}e^{-y}[/tex]

this diff. eq. can be solved with separation of variables.So:

[tex]e^{y}dy=3x^{2}dx[/tex] now integrate both sides and solve for y.
 
So, y = ln(x^3+C)?
 
1,631
4
Thanks stupidmath!!! Now, if I wanted to solve, for example y(0)=1

ln (C) = y - ln(x^3)
ln (C) = 1 - 0
C = e^1 ?

y = ln(x^3) + e^1 ?
 
1,631
4
Well, to impose the initial condition y(0)=1, you actually here have that when x=0, y=1.

So it means:

[tex]1=ln(c)=>C=e[/tex]

so our particular solution would be

[tex]y=ln(x^{3}+e)[/tex]

You just misplaced the parentheses on your final answer.
 
Ah, thanks a bunch stupidmath!!!
 
One last question stupidmath. wouldn't dy/dx = 4(2y-1) end up being y = Ce^-32x?
 
HallsofIvy
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It is not "stupidmath". Beware of his ire!

You should be able to do this one yourself now. If dy/dx= 4(2y-1) then dy/(2y-1)= 4dx.
Integrating on both sides (1/2)ln|2y-1|= 4x+ C so ln|2y-1|= 8x+ C' (C'= 2C). Then 2y-1= C"e^(8c).
 
arildno
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It is suttypud-math, not stupidmath!
 
Oh, I apologize for the faux pas sutupidmath. I also appreciate the help, thanks.
 
It is suttypud-math, not stupidmath!
Is there any meaning to this?:smile:
 
Mute
Homework Helper
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Thanks stupidmath!!! Now, if I wanted to solve, for example y(0)=1

ln (C) = y - ln(x^3)
ln (C) = 1 - 0
C = e^1 ?

y = ln(x^3) + e^1 ?

Remember that ln(a + b) =/= ln(a) + ln(b), which is what you did here.
 

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